a metal sample weighing 44.625 grams and at a temperature of 100.00 C was placed in 35.278 grams of water contained in a calorimeter at 24.74 C. at equilibrium, the temperature of the water and metal was 34.97 C. what is the specific heat of the metal ? what is the approximate atomic weight of the metal ?
specific heat = ________________
atomic weight = _______________
specific heat = ________________
atomic weight = _______________
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you write the energy lost by the matal (Em) is given to the water(Ew)
formula E=mc(tf-ti)
Em = 44.625*c(m) *(34.97-100)= -2901.96*c(m)
Ew= 35.278*4.184*(34.97-24.74)=1509.98 J
you write that the absolute values for Em and Ew are equal
so c(m) = 1509.98/2901.96=0.52J/°K/g
The metal is titanium Ti (see my link) atomic mass 47.857
formula E=mc(tf-ti)
Em = 44.625*c(m) *(34.97-100)= -2901.96*c(m)
Ew= 35.278*4.184*(34.97-24.74)=1509.98 J
you write that the absolute values for Em and Ew are equal
so c(m) = 1509.98/2901.96=0.52J/°K/g
The metal is titanium Ti (see my link) atomic mass 47.857